Nonlinear finite elements/Stress and strain in one and two dimension

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[edit] Stresses and strain in one dimension

[edit] 1-D Strain Measures

  • Engineering strain:
\varepsilon_{E} := \cfrac{l-L}{L} = \cfrac{\Delta L}{L}
  • Natural/Logarithmic/True strain:
\varepsilon_{L} := \int_L^l \cfrac{dl}{l} = \ln(l)\Big|_L^l = \ln\left(\cfrac{l}{L}\right)

Relation between engineering and true strain:

\varepsilon_{L} = \ln\left(\cfrac{L+\Delta L}{L}\right) = \ln\left(1+\cfrac{\Delta L}{L}\right) = \ln\left(1+\varepsilon_{E}\right)
  • Green (Lagrangian) strain:
\varepsilon_{G} := \frac{1}{2}\left(\cfrac{l^2-L^2}{L^2}\right) =\frac{1}{2}\left(\cfrac{l^2}{L^2}-1\right)
  • Almansi-Hamel (Eulerian) strain:
\varepsilon_{A} := \frac{1}{2}\left(\cfrac{l^2-L^2}{l^2}\right) =\frac{1}{2}\left(1-\cfrac{L^2}{l^2}\right)

[edit] 1-D Stress Measures

  • Engineering/Nominal stress:
P = \sigma_{E} := \cfrac{T}{A}
  • Cauchy/True stress:
\sigma = \sigma_{T} := \cfrac{T}{a}

Relation between engineering and true stress (no volume change):

\sigma_{T} = \cfrac{T}{a} = \cfrac{Tl}{AL} = \sigma_{E}\left(\cfrac{L+\Delta L}{L}\right) = \sigma_{E}\left(1 +\varepsilon_{E}\right)

[edit] 1-D Stress-Strain Relations

  • True stress - Green strain:
\sigma_{T} = E_{TG} \left(\cfrac{l^2 - L^2}{2 L^2}\right)
  • True stress - True strain:
\sigma_{T} = E_{TT} \ln\left(\cfrac{l}{L}\right)

[edit] Example

Stretching of a bar
l^2 = D^2 + x^2 \implies \cfrac{dl}{dx} = \cfrac{x}{l}
\sin\theta = \cfrac{x}{l(x)}
  • Assume incompressible material.
V = v ~;~~ AL = al or : V = al~;~~ a = V/l}
  • Equilibrium.
T(x) = F:\implies~~ R(x) = T(x) - F = 0

where

\begin{align}T(x) & = \sigma(x)~a(x)~\sin\theta \\ & = \cfrac{\sigma(x)~a(x)~x}{l(x)} \\ & = {\cfrac{\sigma(x)~V~x}{l(x)^2}}\end{align}

[edit] Stress-Strain relation 1

\sigma(x) = E \left(\cfrac{l(x)^2 - L^2}{2 L^2}\right)

Then,

\begin{align} T(x) & = \cfrac{E~a(x)~x}{l(x)} \left(\cfrac{l(x)^2 - L^2}{2 L^2}\right)\\ & = \cfrac{E~V~x}{l^2} \left(\cfrac{l^2 - L^2}{2 L^2}\right) \end{align}

and

R(x) =\cfrac{E~V~x}{l^2} \left(\cfrac{l^2 - L^2}{2 L^2}\right) - F

Highly nonlinear in x.

[edit] Stiffness

Stiffness = change in equilibrium equation due to change in position.

K(x) = \cfrac{dR(x)}{dx} = \cfrac{dT(x)}{dx} ~~(\text{if}~ F ~\text{ is constant})

Now,

T(x) = \cfrac{\sigma(x)~V~x}{l(x)^2}

Therefore,

\begin{align} \cfrac{dT}{dx} & = V\left[x~\cfrac{d}{dx}\left(\cfrac{\sigma}{l^2}\right) + \cfrac{\sigma}{l^2}\right] \\ & = V\left[x~\cfrac{d}{d\sigma}\left(\cfrac{\sigma}{l^2}\right) \cfrac{d\sigma}{dx} + x~\cfrac{d}{dl}\left(\cfrac{\sigma}{l^2}\right) \cfrac{dl}{dx} + \cfrac{\sigma}{l^2}\right] = V\left[\cfrac{x}{l^2}\cfrac{d\sigma}{dl}\cfrac{dl}{dx} - \cfrac{2x\sigma}{l^3}\cfrac{dl}{dx} + \cfrac{\sigma}{l^2}\right] \\ & = \cfrac{Vx}{l^2}\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{dl}{dx} + \cfrac{V\sigma}{l^2} = \cfrac{Vx}{l^2}\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x}{l} + \cfrac{V\sigma}{l^2} \\ \implies K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \end{align}
\sigma= E \left(\cfrac{l^2 - L^2}{2 L^2}\right) \implies \cfrac{d\sigma}{dl}= \cfrac{E~l}{L^2}
\begin{align} K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = a\left(\cfrac{E~l}{L^2} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = \cfrac{A}{L}\left(E - 2S\right)\cfrac{x^2}{l^2} + \cfrac{S~A}{L} \end{align}

Initial stress/Geometric stiffness

\cfrac{S~A}{L} ~;~~ S= \sigma \cfrac{L^2}{l^2} = P \cfrac{L}{l}

[edit] Stress-Strain relation 2

\sigma(x) = E \ln\left(\cfrac{l(x)}{L}\right)

Then,

\begin{align} T(x) & = \cfrac{E~a(x)~x}{l(x)} \ln\left(\cfrac{l(x)}{L}\right) \\ & = \cfrac{E~V~x}{l^2} \ln\left(\cfrac{l}{L}\right) \end{align}

and

R(x) =\cfrac{E~V~x}{l^2} \ln\left(\cfrac{l}{L}\right) - F

Highly nonlinear in x.

\sigma= E \ln\left(\cfrac{l}{L}\right) \implies \cfrac{d\sigma}{dl}= \cfrac{E}{l}
[edit] Stiffness
\begin{align} K & = a\left(\cfrac{d\sigma}{dl} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = a\left(\cfrac{E}{l} - \cfrac{2\sigma}{l}\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \\ & = \cfrac{a}{l}\left(E - 2\sigma\right)\cfrac{x^2}{l^2} + \cfrac{\sigma~a}{l} \end{align}

Initial stress/Geometric stiffness:

\cfrac{\sigma~a}{l}

[edit] Strain Measures in two dimensions

Strains in two dimensions

[edit] Small strains

\begin{align} \varepsilon_{xx} & = \frac{\partial u_x}{\partial x} \\ \varepsilon_{yy} & = \frac{\partial u_y}{\partial y} \\ \varepsilon_{xy} & = \frac{1}{2}\left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right) \end{align}

For 90o rotation,

u_x = -Y - X ~;~~ u_y = X - Y

Then strains are:

\begin{align} \varepsilon_{xx} & = -1 \\ \varepsilon_{yy} & = -1 \\ \varepsilon_{xy} & = 0 \end{align}

Rotation should not lead to non-zero strains!


[edit] Finite strains

Green strain in two dimensions

For 90o rotation,

u_x = -Y - X ~;~~ u_y = X - Y

Then,

E_{xx} = 0~;~~ E_{yy} = 0~;~~E_{xy} = 0


[edit] Green strain (1-D)

\varepsilon_{G} = \cfrac{l^2-L^2}{2L^2}

In 2-D:

E_{xx} = \cfrac{ds^2-dX^2}{2dX^2}

Now,

ds^2= dX^2\left(1 +\frac{\partial u_x}{\partial X}\right)^2 + dX^2\left(\frac{\partial u_y}{\partial X}\right)^2

Therefore,

\begin{align} E_{xx} & = \cfrac{dX^2}{2dX^2}\left[ \left(1 +\frac{\partial u_x}{\partial X}\right)^2 + \left(\frac{\partial u_y}{\partial X}\right)^2 - 1 \right] \\ & = \frac{\partial u_x}{\partial X} + \frac{1}{2}\left[ \left(\frac{\partial u_x}{\partial X}\right)^2 + \left(\frac{\partial u_y}{\partial X}\right)^2 \right] \end{align}

Similar for Eyy and Exy.

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